19 SUPPORT OF A MEASURE. MEASURES WITH COMPACT SUPPORT 213 (ii) If H is any majorized family of positive measures and v = sup H (13.15.4), then Supp(v) is the closure of the union of the supports of the measures /ieH. This follows immediately from (13.16.1) and (13.15.9), applied to the characteristic function of a relatively compact open set (having regard to (13.15.4)). The complement of Supp(/j) can also be defined as the largest of the interiors of /^-negligible sets. Hence Supp(//) is the intersection of the closures of all sets on which the measure /x is concentrated (13.18). But it should be realized that two disjoint measures can have the same support: for example, an atomic measure on R, concentrated on a denumerable dense set, and such that each point of this set has nonzero measure (13.18.8), has the same support as Lebesgue measure. (13.19.3) For a measure ju on X to be such that every continuous complex- valued function on X is n-integrable, it is necessary and sufficient that SuppGu) should be compact. The mapping f\—> /*(/) is then a continuous linear form on the Fr&chet space #C(X) (12.14.6), and conversely every continuous linear form on «c(X) is of this type. If ju has compact support S, then X — S is ju-negligible, hence/<ps is equal to /almost everywhere. Sincefys is measurable (13.9.6) and \f<ps\ is bounded above by a multiple of (ps (3.17.10), it follows that/ips, and hence also/, is integrable. Let us show conversely that if S = Supp(/z) is not compact, then there exist continuous real-valued functions /^O such that \/i\*(f) = +00. By hypothesis (3.18.3) there exists an increasing sequence of relatively compact open sets Un in X such that Un c Un+1 and the Un cover X. Since X is not compact, we have Un ^ X for all n. We now define inductively a sequence (an) of points of X and a sequence (Vn) of relatively compact open sets, as follows: #! e S; Y! is a neighborhood ofai;an+l belongs to the intersection of S and the complement of the union of On and the Vk with k ^ n (by hypothesis, this intersection is not empty); VB+1 is a relatively compact neighborhood of an+l such that Vn+1 does not intersect 0,, nor any V* with k ^ n, For each n9 let gn be a continuous mapping of X into [0, 1] which takes the value 1 at an and the value 0 throughout X — Vn (4.5.2). The hypothesis implies that |Ju|(^n)>0. Put fn=9nl\V\(9^ so that |^|(/„) = 1. Then the function 00 /= y fn is everywhere finite and continuous (because every point M=l HUNT LIBflftBY rlMfffilE-MELLWI IBHage of /x under TT is Lebesgue measure A on I.